NCERT Solutions For Class 10 Maths Chapter 7

NCERT Solutions Class 10 Maths Coordinate Geometry

NCERT Solutions For Class 10 Maths Chapter 7 Coordinate Geometry are provided here in a detailed and easy way. In this chapter, students are introduced with various important topics on coordinate geometry class 10 formulas and students learn how to calculate the distance between any two points if their coordinates are given, finding the area of a triangle whose three points are given, etc.

Coordinate geometry class 10 questions with NCERT solutions given here are very simple and are extremely easy to understand and any students can clear their doubts instantly. These given solutions for coordinate geometry will help the students to clear their doubts and have a proper understanding of the concepts taught in this chapter. Exercise-wise PDFs is also available here which the students can download and access all the solutions in offline mode.

NCERT Solutions for Class 10 Maths Chapter 7 Exercises

Check the exercise-wise coordinate geometry NCERT solutions for grade 10 below:

1. If the point A (x , y) is equidistant from B (4 , 2) and C (-2 , 4). Find the relation between x and y.

|AB| = $\sqrt{(x-5)^{2}+(y-1)^{2}}$

= $\sqrt{x^{2}+16-8x+y^{2}+4-4y}$

|AC| = $\sqrt{(x+2)^{2}+(y-5)^{2}}$

= $\sqrt{x^{2}+4+4x+y^{2}+16-8y}$

Since, |AB| = |AC|

$\Rightarrow$ $x^{2}+y^{2}-8x-4y+20$ = $x^{2}+y^{2}+4x-8y+20$

$\Rightarrow$      8y – 4y = 4x + 8x

$\Rightarrow$      y = 3x

2. Find the perimeter of a triangle with vertices (0, 8), (0, 0) and (6, 0)

Let the vertices of the $\bigtriangleup$ be P(0, 8), Q(0, 0) and R(6, 0)

∴ PQ = $\sqrt{(0)^{2}+(-8)^{2}}=\sqrt{64}= 8$

∴ QR = $\sqrt{(6)^{2}+(0)^{2}}=\sqrt{36}= 6$

∴ RP = $\sqrt{(-6)^{2}+(8)^{2}}=\sqrt{100}= 10$

∴  Perimeter of $\bigtriangleup$ = 8 + 6 + 10 = 24 units

3.Find the coordinates of the perpendicular bisector of the line segment joining the points P (1, 4) and Q (2, 3), cuts the y-axis.

Here, O (0, y), P (1, 4) and Q (2, 3)

AO = BO

$\Rightarrow$ $\sqrt{(0-1)^{2}+(y-4)^{2}}$ = $\sqrt{(0-2)^{2}+(y-3)^{2}}$

$\Rightarrow$ $1+y^{2}+16-8y= 4+y^{2}+9-6y$

$\Rightarrow$   8y – 6y = 17 -13

$\Rightarrow$   2y = 4

y = 2

∴The required point is (0, 2)

4. The points which divides the line segment joining the points (5, 8) and (1, 2) in ratio 1:2 internally lies in which quadrant?

Now C$\left ( \frac{1+10}{3}, \frac{2+16}{3} \right )$

∴ C$\left ( \frac{11}{3},6 \right )$

Since, C$\left ( \frac{11}{3},6 \right )$ lies in IV quadrant.

5. If A $\left ( \frac{x}{2},3 \right )$ is the mid-point of the line segment joining the points B(4,1) and C(8,7). Find the value of x.

∴$\frac{4+8}{2}=\frac{a}{2}$

$\Rightarrow$ 24 = 2a

$\Rightarrow$ a = 12

6. A line intersects the y-axis at point A and B respectively. If (3, 4) is the mid-point of AB. Find the coordinates of A and B.

Here, $\frac{x+0}{2}=3$ $\Rightarrow x= 6$

$\frac{0+y}{2}=4$ $\Rightarrow x= 8$

∴ The coordinates of A and B are (0, 8) and (6, 0).

7. Find the fourth vertex S of a parallelogram PQRS whose three vertices are P (-3, 8), Q (5, 4) and R (6, 2).

Here, $\frac{-3+6}{2}=\frac{5+x}{2}\Rightarrow x=-1$

$\frac{8+2}{2}=\frac{4+x}{2}\Rightarrow x=3$

∴The fourth vertices S of a parallelogram PQRS is S(-1, 3)

8. Find the coordinates of the point which is equidistant from the three vertices of the

$\bigtriangleup$POQ.

Point equidistant from the three vertices of  a right angle triangle is the mid-point of hypotenuse.

$\Rightarrow \left ( \frac{4x+0}{2},\frac{0+4y}{2} \right )\Rightarrow (2x,2y)$

9. Find the area of a triangle with vertices (x, y + z), (y, z + x) and (z, x + y).

Area of the required

$\bigtriangleup = \frac{1}{2}|x(z+x-x-y)+y(x+y-y-z) +(y+z-z-x)|$

= $\frac{1}{2}|x(z-y)+y(x-z)+z(y-a)|$

= $\frac{1}{2}|0|$ = 0.

10. Find the area of a triangle with vertices P (2, 0), Q (6, 0) and R (5, 4)

Area of the required $\bigtriangleup PQR$

= $\frac{1}{2}|2(0 – 2)+ 6(4 – 0)+ 5(0 – 0)|$

= $\frac{1}{2}|24-4|=10$ Sq. units.

11. If the points O (0, 0), P (3, 4), Q (x, y) are collinear, then write the relation between x and y.

Since, O (0, 0), P (3, 4), Q (x, y) are collinear.

$\Rightarrow$ Area of a triangle formed by these points vanishes

∴ $\frac{1}{2}|0(4 – y)+3(y-0)+x(0-4)|=0$

$\Rightarrow y-2x=0\Rightarrow 2x=y$

12. Find the coordinates of the point O dividing the line segment joining the point P (6, 5)

and Q(2, 10) in the ratio 2:1.

$x=\frac{2(2)+1(6)}{5}= \frac{4+6}{5}= 2$

and $y=\frac{2(10)+1(5)}{5}= \frac{20+5}{5}= 5$

∴ O(2, 5)

13. The line joining A(4, 3) and B(-3, 6) meets y-axis at C. At what ratio dose C divides the line segment AB?

Since R lies on the y-axis, let C be (0, y)

Suppose that the required ratio be z:1

∴$x=\frac{-3z+4}{z+1}\Rightarrow 0= -3z+4\Rightarrow z=\frac{4}{3}$

∴ The required ratio is 4 : 3.

In coordinate geometry class 10 questions with solutions, students are introduced to several important concepts like Coordinates of a point, Parallel and Perpendicular Lines, Distance Formula & Section formula, Area of triangle, Locus of a point, Transformation of axis, Concepts of straight line, Conic Section (Parabola & Hyperbola, Circle, Ellipse, Pair of straight lines) and more. These concepts are extremely important for CBSE class 10 board exam as almost 6 marks are included in the exam from this chapter.

All these concepts are extremely crucial as questions are asked in the 10th class competitive exams. Apart from that, these concepts are also included in some of the important maths topics in the later grades. Students are required to be thorough with all the topics that are included in this chapter 7 and should understand the NCERT solved examples properly.

Students also need to solve exercise problems in this chapter and can refer to this NCERT Solutions for Class 10 Maths Coordinate Geometry in case of any doubts. Apart from this, students can also find the solutions for all the chapters by visiting NCERT Solutions For Class 10 Maths. Keep visiting BYJU’S to get complete assistance for CBSE class 10 board exams. Students are provided with sample papers, question papers, and notes to help them to prepare for their exams in a more effective way. Students are also suggested to download BYJU’S- The Learning App to get a personalized and engaging learning experience.